[SS] Ch04 Ex : Inverse Laplace Transform

1. 다음 $X(s)$의 Inverse Laplace Transform을 구하라.

$$X(s)=\frac{8s^2-7s-6}{s^3-s^2-6s}$$ sol. $$\begin {aligned} X(s)&=\frac{8s^2-7s-6}{s^3-s^2-6s}\\ &=\frac{8s^2-7s-6}{s(s-3)(s+2)}\\ &=\frac{A}{s}+\frac{B}{s-3}+\frac{C}{s+2} \end {aligned}$$

 

distinct pole의 경우로서 다음과 같이 A,B,C를 구할 수 있음.

$$ \begin {aligned} A&= \left. \frac{8s^2-7s-6}{(s-3)(s+2)} \right |_{s=0} \\ &= \frac{0-0-6}{-3\cdot 2} \\ &= 1 \end {aligned} $$

$$ \begin {aligned} B&= \left. \frac{8s^2-7s-6}{(s)(s+2)} \right |_{s=3} \\ &= \frac{72-21-6}{3\cdot 5} = \frac{45}{15} \\ &= 3 \end {aligned} $$

$$ \begin {aligned} C&= \left. \frac{8s^2-7s-6}{(s)(s-3)} \right |_{s=-2} \\ &= \frac{32+14-6}{-2\cdot -5} = \frac{40}{10} \\ &= 4 \end {aligned} $$

 

즉, $X(s)$ 는 다음과 같음.

$$X(s)=\frac{1}{s}+\frac{3}{s-3}+\frac{4}{s+2}$$

$x(t)$는 다음과 같음. (Laplace Transform Table참고)

$$x(t) = u(t) + 3e^{-(-3)t}u(t) + 4e^{-2t}u(t)$$

ROC는 다음과 같음. $$Re(s)>0 \\ Re(s)>-(-3)=3 \\ Re(s)>-2\\ \therefore Re(s)>3$$

2. 다음 $X(s)$의 Inverse Laplace Transform을 구하라.

$$X(s)=\frac{8s+10}{(s+1)(s+2)^3}$$ sol. $$ \begin {aligned} X(s) &= \frac{8s+10}{(s+1)(s+2)^3} \\ &= \frac{A}{(s+1)}+\frac{B_1}{(s+2)}+\frac{B_2}{(s+2)^2}+\frac{B_3}{(s+2)^3}\end{aligned} $$

 

$A$와 $B_3$는 distinct poles의 경우와 마찬가지로 구할 수 있음.

$$ \begin {aligned} A &= \left . X(s)(s+1) \right |_{s=-1} \\ &= \left . \frac{8s+10}{(s+1)(s+2)^3}(s+1) \right |_{s=-1} \\ &= \frac {2}{1^3} \\ &= 2 \end {aligned}$$

$$ \begin {aligned} B_3 &= \left . X(s)(s+2)^3 \right |_{s=-2} \\ &= \left . \frac{8s+10}{(s+1)(s+2)^3}(s+2)^3 \right |_{s=-2} \\ &= \left . \frac{8s+10}{(s+1)} \right |_{s=-2} \\ &= \frac {8\cdot-2+10}{(-1)} \\ &= 6 \end {aligned}$$

 

$B_2$는 다음과 같이 한번 미분을 수행하여 구함.

$$ \begin {aligned} B_2 &= \left . \frac{ \text{d}}{\text{d}s} \left[ X(s)(s+2)^3\right] \right |_{s=-2} \\ &= \left . \frac{8(s+1)-(8s+10)\cdot1}{(s+1)^2} \right |_{s=-2} \\ &= \frac {-2}{1} \\ &= -2 \end {aligned}$$

 

$B_1$은 다음과 같이 구함.

$$ \begin {aligned} B_1 &= \left . \frac{1}{2}\frac{ \text{d}^2}{\text{d}s^2} \left[ X(s)(s+2)^3\right] \right |_{s=-2} \\ &= \frac{1}{2}\left . \frac{ \text{d}}{\text{d}s}\frac{8(s+1)-(8s+10)\cdot1}{(s+1)^2} \right |_{s=-2} \\ &= \frac{1}{2}\left . \frac{ \text{d}}{\text{d}s}\frac{-2}{(s+1)^2} \right |_{s=-2} \\ &= \frac{1}{2} \left .\frac {-2\cdot-2}{(s+1)^3} \right |_{s=-2}\\ &= \left .\frac {2}{(s+1)^3} \right |_{s=-2}\\ &= -2 \end {aligned}$$

 

이를 통해, $X(s)$, $x(t)$는 다음과 같음.

$$ \begin {aligned} X(s) &= \frac{8s+10}{(s+1)(s+2)^3} \\ &= \frac{A}{(s+1)}+\frac{B_1}{(s+2)}+\frac{B_2}{(s+2)^2}+\frac{B_3}{(s+2)^3}\\ &= \frac{2}{(s+1)}+\frac{-2}{(s+2)}+\frac{-2}{(s+2)^2}+\frac{6}{(s+2)^3} \\ x(t)&= 2e^{-t} u(t)-2e^{-2t}u(t) -2te^{-2t}u(t)+6\cdot\frac{1}{2}t^2 e^{-2t}u(t) \\ &= 2e^{-t} u(t)-2e^{-2t}u(t) -2te^{-2t}u(t)+3t^2 e^{-2t}u(t) \end {aligned}$$